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Lasse Rempe
Lasse Rempe
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The answer to your question is negative.

EDIT I have to be brief, so I will only give an outline. I will try to expand on it tomorrowadded some additional details and made some corrections.

In the entire case, it is possible to construct an entire function with the following propertyproperties: (a) The Fatou set consists of a single conneted attracting basin basin; (henceb) If $C$ is a component of the nonescapingJulia set is connected), then the set of non-escaping part of the Julia set contains no curvespoints in $C$ is totally disconnected (actually,in fact has Hausdorff dimension zero), and there (c) There is a component of $J(f)$ that contains a non-escaping point in the Julia set, but no point that is not accessible from $F(f)$.

Now, by (a), the nonescaping set is connected (since it contains a dense connected subset of the plane). On the other hand, it can be shown that there is no curve connecting the non-escaping points in (c) to a point in the Fatou set without intersecting the escaping set.

This shows thatHence the non-escaping set is not path-wise connected, and hence not contractible. (The construction is contained in an upcoming article of mine, dealing more generally with the topology of transcendental Julia sets.)

For quadratic Cremer polynomials, the key point is that the Cremer point $z_0$ is accumulated on by small cycles by work of Yoccoz. Now, and some ofif the arcsJulia set is path-connected (if anyotherwise, there is nothing to prove), then there is a unique arc connecting each of these periodic points to $z_0$.

Now, for any cycle, it follows from the fixed pointwork of Perez-Marco that at least one of the corresponding arcs has diameter at least $\delta$, for $\delta$ independent of the cycle. Indeed, I believe it follows that at least one of them must be quite longcontain the critical point.

From this, one can deduce (although I haven't made sure to check all the details) that the Julia set is not contractible.

The answer to your question is negative. I have to be brief, so I will only give an outline. I will try to expand on it tomorrow.

In the entire case, it is possible to construct an entire function with the following property: The Fatou set consists of a single conneted attracting basin (hence the nonescaping set is connected), the non-escaping part of the Julia set contains no curves (actually, has Hausdorff dimension zero), and there is a non-escaping point in the Julia set that is not accessible from the Fatou set.

This shows that the non-escaping set is not path-wise connected, and hence not contractible. (The construction is contained in an upcoming article of mine, dealing more generally with the topology of transcendental Julia sets.)

For quadratic Cremer polynomials, the key point is that the Cremer point is accumulated on by small cycles by work of Yoccoz, and some of the arcs (if any) connecting these to the fixed point must be quite long.

The answer to your question is negative.

EDIT I have added some additional details and made some corrections.

In the entire case, it is possible to construct an entire function with the following properties: (a) The Fatou set consists of a single conneted attracting basin; (b) If $C$ is a component of the Julia set, then the set of non-escaping points in $C$ is totally disconnected (in fact has Hausdorff dimension zero), (c) There is a component of $J(f)$ that contains a non-escaping point, but no point that is accessible from $F(f)$.

Now, by (a), the nonescaping set is connected (since it contains a dense connected subset of the plane). On the other hand, it can be shown that there is no curve connecting the non-escaping points in (c) to a point in the Fatou set without intersecting the escaping set.

Hence the non-escaping set is not path-wise connected, and hence not contractible. (The construction is contained in an upcoming article of mine, dealing more generally with the topology of transcendental Julia sets.)

For quadratic Cremer polynomials, the key point is that the Cremer point $z_0$ is accumulated on by small cycles by work of Yoccoz. Now, if the Julia set is path-connected (otherwise, there is nothing to prove), then there is a unique arc connecting each of these periodic points to $z_0$.

Now, for any cycle, it follows from the work of Perez-Marco that at least one of the corresponding arcs has diameter at least $\delta$, for $\delta$ independent of the cycle. Indeed, I believe it follows that at least one of them must contain the critical point.

From this, one can deduce (although I haven't made sure to check all the details) that the Julia set is not contractible.

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Lasse Rempe
Lasse Rempe
  • 6.5k
  • 1
  • 28
  • 47

The answer to your question is negative. I have to be brief, so I will only give an outline. I will try to expand on it tomorrow.

In the entire case, it is possible to construct an entire function with the following property: The Fatou set consists of a single conneted attracting basin (hence the nonescaping set is connected), the non-escaping part of the Julia set contains no curves (actually, has Hausdorff dimension zero), and there is a non-escaping point in the Julia set that is not accessible from the Fatou set.

This shows that the non-escaping set is not path-wise connected, and hence not contractible. (The construction is contained in an upcoming article of mine, dealing more generally with the topology of transcendental Julia sets.)

For quadratic Cremer polynomials, the key point is that the Cremer point is accumulated on by small cycles by work of Yoccoz, and some of the arcs (if any) connecting these to the fixed point must be quite long.